Question

Is it possible to rewrite x(-1)^x-(-1)^x=y(-1)^y-(-1)^y

In such a way i could reduce it to x=y ?

Answer 1

\[x,y \ge0\]

Answer 2

is that?
\[x(-1)^x-(-1)^x=y(-1)^y-(-1)^y\]
?

Answer 3

I'd factor out \((-1)^x\) and \((-1)^y\) first

Answer 4

Because if you did that first, then you'd have

\((-1)^x(x - 1) = (-1)^y(y - 1)\)

Answer 5

Afterwards, you can apply all the stuff @myininaya was talking about

Answer 6

After applying all those steps, you should end with the desired result.

Answer 7

thats where i started from

Answer 8

Can you show all the steps you have completed up to this point?

Answer 9

\[(-1)^x(x-1)=(-1)^y(y-1)\] that is

Answer 10

sure

Answer 11

Apparently only two of the four cases presented by @myininaya actually work.

Answer 12

Proving injection by contrapositive method
\[f(x)=f(y)\]
\[\frac{ ((-1)^x(2x-1)+1) }{ 4 }=\frac{ ((-1)^y(2y-1)+1) }{ 4 }\]
.
.
.
.
\[(-1)^x(x-1)=(-1)(y-1)\]
is as far as i've come.
I don't know if there's another way to prove it with sets f: N -> Z

Answer 13

multiplied both sides with 4
subtracted 1 on both sides
expaneded the brackets
divided each expression with 2

Answer 14

ect...

Answer 15

etc*

Answer 16

How did you conclude that \( ((-1)^x(2x-1)+1)=((-1)^y(2y-1)+1)\) is equivalent to \((-1)^x(x-1)=(-1)(y-1)\)

What did you do with the 2?

Answer 17

\[2x(-1)^x-(-1)^x=2y(-1)^y-(-1)^y\]
\[<->\]
\[\frac{ 2y(-1)^x-(-1)^x }{ 2 }=\frac{ 2y(-1)^y-(-1)^y }{ 2 }\]

Answer 18

ah i see hehe

Answer 19

\[x(-1)^x-(-1)^x=y(-1)^y-(-1)^y\]
<->
\[(-1)^x(x-1)=(-1)^y(y-1)\]

Answer 20

can't see where i should go from here : / (-1)^y,x is annoying me

Answer 21

Based on what you posted originally, you could simplify it to

\((-1)^x(2x - 1) = (-1)^y(2x - 1)\)

I don't know how you got it to the other form you have. What you have is not an eqivalent form of the original equation.

Answer 22

Okay, i stil don't have an idea on what to do, from that expression.
\[(-1)^x(2x-1)=(-1)^y(2y-1)\]

Answer 23

atm i can't see how i should simplify it, without disturbing either one of the sides

Answer 24

Well I deleted my cases because I didn't know if x and y were actually integers.

Answer 25

ye, the original function maps from N to Z

Answer 26

go with applying the cases
like hero said 2 of the 4 will work
those cases being both odd or both even

Answer 27

of course you won't have the function is invective if you assume one is even and one is odd because that would imply you don't have
f(a)=f(b) implies a=b

because there is no way an even can equal an odd

Answer 28

injective*

Answer 29

I see, i think i have to prove it in another way

Answer 30

but thank you very much :)!